Solve $u_{tt}=u_{xx}$ , $u(x,0)=1-2x+3x^2$, $t\gt0,-\infty\lt x\lt +\infty$

Question

Solve $u_{tt}=u_{xx}$ , $u(x,0)=1-2x+3x^2$, $t\gt0,-\infty\lt x\lt +\infty$

For $u_{tt}=u_{xx}$ ,we can get $u(x,t)=f(x-t)+g(x+t)$ and $u(x,0)=1-2x+3x^2=f(x)+g(x)$, what's the next? Now do we need more information to get an explict solution?

Answer 1

$$ u(x,t) = f(x-t)+g(x+t) $$

$$ u(x,0) = f(x)+g(x) = 1-2x+3x^2 $$

so making $f = g$ we have

$$ f(x) = \frac 12\left(1-2x+3x^2\right) $$

and finally

$$ u(x,t) = \frac 12\left(1-2(x-t)+3(x-t)^2\right) + \frac 12\left(1-2(x+t)+3(x+t)^2\right) $$

or

$$ u(x,t) = \frac 12\left(1+3t^2+x(3x-2)\right) $$